# Question ef7d5

Oct 18, 2016

The answer is the option $\left(2\right)$

#### Explanation:

The parallelepiped volume is calculated by doing

$V = \left\mid \left\langle\vec{a} \times \vec{b} , \vec{c}\right\rangle \right\mid$ so instead

$V ' = \left\mid \left\langle\left(\vec{b} + \vec{c}\right) \times \left(\vec{c} + \vec{a}\right) , \vec{a} + \vec{b}\right\rangle \right\mid$

Expanding

$\left(\vec{b} + \vec{c}\right) \times \left(\vec{c} + \vec{a}\right) = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a}$

and

$\left\langle\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a} , \vec{a} + \vec{b}\right\rangle =$

$= \left\langle\vec{b} \times \vec{c} , \vec{a}\right\rangle + \left\langle\vec{c} \times \vec{a} , \vec{b}\right\rangle = 2 \left\langle\vec{b} \times \vec{c} , \vec{a}\right\rangle$

so finally

$V ' = 2 V$

Oct 18, 2016

2) $80$

#### Explanation:

We will use several facts.

First, note that the volume of a parallelepiped with coterminous edges $\vec{a} , \vec{b} , \vec{c}$ can be found by the triple scalar product$\vec{a} \cdot \left(\vec{b} \times \vec{c}\right)$. As this may be a signed value, we will take the absolute value to get the volume. This allows us to interpret the given information as $| \vec{a} \cdot \left(\vec{b} \times \vec{c}\right) | = 40$.

Two more facts about the scalar triple product we will use:

• $\vec{a} \cdot \left(\vec{b} \times \vec{c}\right) = | \left({a}_{1} , {a}_{2} , {a}_{3}\right) , \left({b}_{1} , {b}_{2} , {b}_{3}\right) , \left({c}_{1} , {c}_{2} , {c}_{3}\right) |$

• $\vec{a} \cdot \left(\vec{b} \times \vec{c}\right) = - \vec{b} \cdot \left(\vec{a} \times \vec{c}\right) = - \vec{c} \cdot \left(\vec{b} \times \vec{a}\right)$

Note that the first fact means that any scalar triple product involving the same vector twice will clearly be $0$, as it is equivalent to taking a determinant of a matrix with repeated rows. Additionally, the cross product of any vector with itself can be immediately evaluated as $\vec{0}$.

Finally, we will use that both the dot product and the cross product distribute over addition. With all that, we will now calculate the volume of the new parallelepiped as the scalar triple product of its coterminous edges:

$| \left(\vec{a} + \vec{b}\right) \cdot \left[\left(\vec{a} + \vec{c}\right) \times \left(\vec{b} + \vec{c}\right)\right] |$

=|(vec(a)+vec(b)) * [(vec(a) xx vec(b)) + (vec(a) xx vec(c)) + (vec(c) xx vec(b)) + cancel((vec(c) xx vec(c))]|#

$= | \cancel{\vec{a} \cdot \left(\vec{a} \times \vec{b}\right)} + \cancel{\vec{a} \cdot \left(\vec{a} \times \vec{c}\right)}$

$+ \vec{a} \cdot \left(\vec{c} \times \vec{b}\right) + \cancel{\vec{b} \cdot \left(\vec{a} \times \vec{b}\right)}$

$+ \vec{b} \cdot \left(\vec{a} \times \vec{c}\right) + \cancel{\vec{b} \cdot \left(\vec{c} \times \vec{b}\right)} |$

$= | - \vec{a} \cdot \left(\vec{b} \times \vec{c}\right) - \vec{a} \cdot \left(\vec{b} \times \vec{c}\right) |$

$= 2 | \vec{a} \cdot \left(\vec{b} \times \vec{c}\right) |$

$= 2 \left(40\right)$

$= 80$